Question

Show that the peak inductor voltage in a series RLC circuit is maximum at frequency ωL=(1ω02−12R2C2)−1/2\omega_L = \left( \frac{1}{\$$omega_0^2$$} - \frac{1}{2} $$R^2$$ $$C^2$$ \$$right)^{-1/2}.$$

Accepted Answer

Start by recalling the resonance condition for a series RLC circuit. The resonance frequency, $$ \omega_0 $$, is given by $$ \omega_0 = \frac{1}{\sqrt{LC}} . $$This is the frequency at which the impedance of the circuit is purely resistive, and the reactive components cancel each other out. The voltage across the inductor, $$ V_L $$, is given by $$ V_L = I \cdot X_L $$, where $$ I  is $$the current in the circuit and $$ X_L = \omega L  is $$the inductive reactance. The current $$ I  in $$the circuit is determined by $$ I = \frac{V}{Z} $$, where $$ Z  is $$the total impedance of the circuit. The total impedance $$ Z  of $$the series RLC circuit is $$ Z = \sqrt{R^2 + (X_L - X_C)^2} $$, where $$ X_C = \frac{1}{\omega C}  is $$the capacitive reactance. Substituting $$ X_L $$ and $$ X_C $$, $$ Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} . To $$maximize the voltage across the inductor, $$ V_L $$, we need to maximize $$ X_L $$ while minimizing $$ Z . $$This occurs at a specific frequency $$ \omega_L . By $$analyzing the derivative of $$ V_L $$ with respect to $$ \omega $$ and setting it to zero, we find that the peak inductor voltage occurs at $$ \omega_L = \left( \frac{1}{\omega_0^2} - \frac{1}{2} R^2 C^2 \right)^{-1/2} . $$Finally, verify the result by substituting $$ \omega_L $$ back into the expressions for $$ X_L $$, $$ Z $$, and $$ V_L  to $$confirm that $$ V_L  is $$indeed maximized at this frequency. This involves algebraic manipulation and simplification to ensure consistency with the given formula.

Show that the peak inductor voltage in a series RLC circuit is maximum at frequency $ω_{L} = {(\frac{1}{ω_{0}^{2}} - \frac{1}{2} R^{2} C^{2})}^{- 1 / 2}$ .

Verified video answer for a similar problem:

Key Concepts

RLC Circuit

Resonant Frequency

Impedance

Show that the peak inductor voltage in a series RLC circuit is maximum at frequency ωL=(1ω02−12R2C2)−1/2\(\omega\)_L = \(\left\)( \(\frac{1}{\omega_0^2}\) - \(\frac{1}{2}\) R^2 C^2 \(\right\))^{-1/2}.

Verified video answer for a similar problem:

Key Concepts

RLC Circuit

Resonant Frequency

Impedance

Show that the peak inductor voltage in a series RLC circuit is maximum at frequency $ω_{L} = {(\frac{1}{ω_{0}^{2}} - \frac{1}{2} R^{2} C^{2})}^{- 1 / 2}$ .